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\topic{Lecture 9 \\Multiple Integral\\ \scriptsize Gamma and  Beta Functions
(1 Oct 2009)
}
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\section{Gamma Function}
The integral 
\begin{equation}
\Gamma(x) = \int_0^{\infty} t^{x-1}e^{-t}dt
\label{Gamma}
\end{equation}
converges for all values $x>0$ and is called Gamma function. It is obvious that integrad is a function of $x$ and $t$, but on integration $t$ gets removed, thus resulting function is function of $x$ only. We denote this function by $\Gamma(x)$. In practice, the integration is posible only for special values of $x$. For other values of $x$, recource must be had to numerical methods for evaluating the integral. However, a number of proprties of the function can be derived from the definition itself.

\section{Important Results}
\subsection{Value of $\Gamma(1)$}
In equation \ref{Gamma} put $x=1$, then solve
\[\Gamma(1) = \int_0^{\infty} e^{-t}dt = \left[-e^{-t} \right]_0^{\infty} = 1 \]
Thus
\begin{equation}
\Gamma(1) = 1
\label{R1}
\end{equation}
Here, it is important that $\Gamma(0)$ is not defined.
\subsection{Relation $\Gamma(x+1) = x \Gamma(x)$}
Consider the integral,
\[\int_0^{\infty} t^xe^{-t} dt\]
On integrating this integral by parts, we see that
\[\int_0^{\infty} t^xe^{-t} dt = \left[-te^{-t}\right]_0^{\infty} + x \int_0^{\infty} t^{x-1}e^{-t}dt \]
The integrated part vanishes as at lower limit, since $x>0$. At the upper limit, we have by L'Hospital's rule
\[\lim_{t\rightarrow \infty} t^x e^{-t} =\lim_{t\rightarrow \infty} \frac{t^x}{e^{t}}=\lim_{t\rightarrow \infty} \frac{xt^{x-1}}{e^{t}}=\lim_{t\rightarrow \infty} \frac{x(x-1)t^{x-2}}{e^{t}}= ... \]
By differentiating the numerator and the denominator again and again, till we get a zero or negative exponent in the numerator, we see that the limit is zero. Hence
\[\int_0^{\infty} t^xe^{-t} dt = x \int_0^{\infty} t^{x-1}e^{-t}dt \]
\begin{equation}
\Gamma(x+1) = x \Gamma(x)
\label{R1}
\end{equation}
\subsection{Relation $\Gamma(n+1) = n!$}
A repeated application of above relation \ref{R1}, we have
\[\Gamma(5) = 4 \Gamma(4) = 4. 3 \Gamma(3) = 4.3.2.\Gamma(2)= 4.3.2.1.\Gamma(1) = 4!  \]
Thus the relation $\Gamma(n+1) = n!$ holds only if $n$ is positive integer. Now if $n$ is not integer, the procedure of evaluation excepet the last value  of $\Gamma(n)$ is taken from the Gamma Function Table; e.g. 
\[\Gamma(4.3) = 3.3 \Gamma(3.3) = 3.3 \times 2.3 \Gamma(2.3) = 3.3 \times 2.3 \times 1.3 \Gamma(1.3)  \]
We generally stop at this stage, and sustitute the value of $\Gamma(1.3)$ from the table of gamma functions.
\section{Extended definition of Gamma function}
The relation \ref{R1} can also be used for defining  the gamma function for negative values of $x$. Thus the complete definition is

\begin{equation}
\begin{array}{ll}
\Gamma(x)& = \int_0^{\infty} t^{x-1}e^{-t}dt,~~~x>0 \\
			& = \frac{\Gamma(x+1)}{x},~~~x<0
\end{array}
\label{GammaE}
\end{equation}
\section*{Problems}
\begin{enumerate}
	\item Evaluate $\displaystyle \int_0^{\infty} \sqrt{x}e^{-\sqrt[3]{x}}dx$.
	\item Evaluate $\displaystyle \int_0^{\infty} \sqrt[4]{x}e^{-\sqrt{x}}dx$.
	\item Evaluate $\displaystyle \int_0^{\infty} x^{n-1}e^{-h^2x^2}dx$.
	\item Evaluate $\displaystyle \int_0^{\infty} \frac{x^a}{a^x}dx$.
	\item Evaluate $\displaystyle \int_0^{\infty} x^{n-1} \left[\ln \left(\frac{1}{x}\right)\right]^{m-1}dx$.
	\item Show that $\displaystyle \int_0^{\infty} e^{-ky}y^{n-1} dy = \frac{\Gamma(n)}{k^n}$.
	\item Show that $\Gamma(1/2) = \sqrt{\pi}$. (REMEMBER THIS RESULT)
	\item Show that $\displaystyle \int_0^{\infty} \left[\ln \left(\frac{1}{x}\right)\right]^{m-1}dx = \Gamma(m)$.
	\item Evaluate (i) $\Gamma(-1/2)$ (ii) $\Gamma(-3/2)$ (iii) $\Gamma(-15/2)$  (iv) $\Gamma(7/2)$ (v) $\Gamma(0)$.
	\item Evaluate $\displaystyle \int_0^{\infty} \farc{dx}{\sqrt{-x \ln x}} $.
	\item Evaluate $\displaystyle \int_0^{\infty}\int_0^{\infty} e^{-(ax^2+by^2)x^{(2m-1)}y^{(2n-1)}}dx dy,\;\;a,b,m,n>0$.
	\iten Show that $1.3.5 \ldots (2n-1) = \frac{2^n\Gamma(n+\frac{1}{2})}{\sqrt{\pi}}$
\end{enumerate}
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